[2^4+1/4)(4^4+1/4)(6^4+1/4)(8^4+1/4)(10^4+1/4)】/[1^4+1/4)(3^4+1/4)(5^4+1/4)(7^4+1/4)(9^4+1/4)]

来源：学生作业帮助网编辑：作业帮时间：2024/04/28 22:51:36

[2^4+1/4)(4^4+1/4)(6^4+1/4)(8^4+1/4)(10^4+1/4)】/[1^4+1/4)(3^4+1/4)(5^4+1/4)(7^4+1/4)(9^4+1/4)]
[2^4+1/4)(4^4+1/4)(6^4+1/4)(8^4+1/4)(10^4+1/4)】/[1^4+1/4)(3^4+1/4)(5^4+1/4)(7^4+1/4)(9^4+1/4)]

[2^4+1/4)(4^4+1/4)(6^4+1/4)(8^4+1/4)(10^4+1/4)】/[1^4+1/4)(3^4+1/4)(5^4+1/4)(7^4+1/4)(9^4+1/4)]
a^4+1/4
=(a^2+1/2)^2-a^2
=(a^2+a+1/2)(a^2-a+1/2)
=[(a+1/2)^2+1/4][(a-1/2)^2+1/4]
根据这个式子,原式可化为：
[2^4+1/4)(4^4+1/4)(6^4+1/4)(8^4+1/4)(10^4+1/4)】/[1^4+1/4)(3^4+1/4)(5^4+1/4)(7^4+1/4)(9^4+1/4)]
={[(2+1/2)^2+1/4][(2-1/2)^2+1/4]*[(3+1/2)^2+1/4][(3-1/2)^2+1/4]*...*[(10+1/2)^2+1/4][(10-1/2)^2+1/4]}/{[(1+1/2)^2+1/4][(1-1/2)^2+1/4]*...*[(9-1/2)^2+1/4][(9+1/2)^2+1/4]}
消去一样的式子：
比如：[(2-1/2)^2+1/4]=[(1+1/2)^2+1/4]
...
所以：
=[(10+1/2)^2+1/4]/[(1-1/2)^2+1/4]
=61
不知道中间的消去有没有错,电脑上看的晕了～
希望我的回答让你满意

孩子别懒，叫计算器来，真不知道它是干嘛吃的？

[2^4+1/4)(4^4+1/4)(6^4+1/4)(8^4+1/4)(10^4+1/4)】/[1^4+1/4)(3^4+1/4)(5^4+1/4)(7^4+1/4)(9^4+1/4)] 1,2,4,6, (（1^4+1/4)(3^4+1/4)(5^4+1/4).(19^4+1/4))/((2^4+1/4)(4^4+1/4)(6^4+1/4).(20^4+1/4)) 求[(1^4+1/4)(3^4+1/4)(5^4+1/4)……(19^4+1/4)]/[(2^4+1/4)(4^4+1/4)(6^4+1/4)……(20^4+1/4)]求[(1^4+1/4)(3^4+1/4)(5^4+1/4)……(19^4+1/4)]/[(2^4+1/4)(4^4+1/4)(6^4+1/4)(20^4+1/4)] 1、2、2、4、3、6、4、( )、( 计算：[1/4（1^4+3^4+5^4+……+19^4)]/[1/4(2^4+4^4+6^4+……+20^4)] 填上合适的运算符号和括号2 2 2 2 2 = 0 2 2 2 2 2 = 14/4+4-4=1 4/4+4/4=2 4 4 4 4=3 4 4 4 4=44 4 4 4=5 4 4 4 4=6 4 4 4 4=7 4 4 4 4=84 4 4 4=9 4 4 4 4=10 1+6+6+8+4+2+6+6+5+4+4+1+4+7+45 1,2,4,4,1,( ) 4 4 4 4=1 4 4 4 4=2 4 4 4 4=3 填标点符号数学(2^4+1/4)(4^4+1/4)(6^4+1/4)(8^4+1/4)(10^4+1/4)/(1^4+1/4)(3^4+1/4)(5^4+1/4)(7^4+1/4)(9^4+1/4)分数线前后是一个整体 1+2+4+6@$456 1/2+1/2+4+1/2+4+6+.+1/2+4+6+...+100 1+1/2+4+1/2+4+6+1/2+4+6+8+.+1/2+4+6+8...+20 已知1^4+2^4+3^4+4^4=354,求2^4+4^4+6^4+8^4=?怎么计算的? 计算 1/2*4+1/4*6+.1/2006*2008 1/2*4+1/4*6+...+1/2012*2014=? 1/2*4+1/4*6+.+1/2008*2010